System impulse response and Convolution by matlab
이전 댓글 표시
Hi everyone, i am very new to matlab, and would like to konw how to obtain y(t) from below
Generate a unit step function as the input function, x(t), and an exponentially decay function as the impulse response function, h(t), such as h(t)=exp(-t/2) (note: 2 is the time constant of the system dynamic response). Using MATLAB to calculate the output of the system, y(t).
Thank you so much in advance
댓글 수: 2
Seikh Rana
2018년 4월 22일
Convolution (random position zero) in matlab plz help me
Seikh Rana
2018년 4월 22일
</matlabcentral/answers/uploaded_files/114156/convooooo.JPG> this picture problem solved in matlab plz help me
답변 (3개)
psyx21
2011년 4월 29일
1 개 추천
hey friends what will be the peak acceleration response for 100g 6ms half sine pulse with zeta=0. I need a matlab script with natural frequency on x axis..thanks
Paulo Silva
2011년 2월 20일
t=0:0.1:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=u.*h;
plot(t,y)
댓글 수: 8
gedaa
2011년 2월 21일
Paulo Silva
2011년 2월 21일
it's the same as
u=zeros(size(t))
It creates a variable with the same size as t but full of zeros. I learned that trick here at matlabcentral :)
David Young
2011년 2월 21일
Paulo - I just wonder if this is right. In the time domain, don't you need to convolve the input and the impulse response rather than multiplying them?
Paulo Silva
2011년 2월 21일
Yes you are correct, try
t=0:0.1:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=conv(u,h);
plot(y)
Paulo Silva
2011년 2월 21일
Correction for the amplitude and time scale
T=0.1;
t=0:T:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=conv(u,h);
plot(t,T*y(1:numel(t)))
gedaa
2011년 2월 21일
Paulo Silva
2011년 2월 21일
I'm not sure about the amplitude at
plot(t,T*y(1:numel(t)))
with the step function is good compared to
step(tf([1],[1 1/2])) but with the impulse it's plot(t,y(1:numel(t))) without the T, I can't figure out why that happens.
gedaa
2011년 2월 23일
Arpan Patel
2021년 2월 25일
0 개 추천
t=0:0.1:10;
u=0*t;
u(t>=0)=1;
h=exp(-t/2);
y=u.*h;
plot(t,y)
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